Php Program To Check If A String Is Substring Of Another

Given two strings s1 and s2, find if s1 is a substring of s2. If yes, return the index of the first occurrence, else return -1.

Examples :

Input: s1 = "for", s2 = "geeksforgeeks"
Output: 5
Explanation:
String "for" is present as a substring
of s2.
Input: s1 = "practice", s2 = "geeksforgeeks"
Output: -1.
Explanation:
There is no occurrence of "practice" in
"geeksforgeeks"

Simple Approach: The idea is to run a loop from start to end and for every index in the given string check whether the sub-string can be formed from that index. This can be done by running a nested loop traversing the given string and in that loop run another loop checking for sub-string from every index. 
For example, consider there to be a string of length N and a substring of length M. Then run a nested loop, where the outer loop runs from 0 to (N-M) and the inner loop from 0 to M. For very index check if the sub-string traversed by the inner loop is the given sub-string or not. 

PHP

<?php
// PHP program to check if a
// string is substring of other.

// Returns true if s1 is substring
// of s2
function isSubstring($s1, $s2)
{
    $M = strlen($s1);
    $N = strlen($s2);

    // A loop to slide
    // pat[] one by one
    for ($i = 0; $i <= $N - $M; $i++)
    {
        $j = 0;

        // For current index i,
        // check for pattern match
        for (; $j < $M; $j++)
            if ($s2[$i + $j] != $s1[$j])
                break;

        if ($j == $M)
            return $i;
    }

    return -1;
}

// Driver Code
$s1 = "for";
$s2 = "geeksforgeeks";
$res = isSubstring($s1, $s2);
if ($res == -1)
    echo "Not present";
else
    echo "Present at index " . $res;

// This code is contributed by mits
?>

Output:

Present at index 5

Complexity Analysis:

• Time complexity: O(m * n) where m and n are lengths of s1 and s2 respectively. 
  A nested loop is used the outer loop runs from 0 to N-M and inner loop from 0 to M so the complexity is O(m*n).
• Space Complexity: O(1). 
  As no extra space is required.

An efficient solution is to use a O(n) searching algorithm like KMP algorithm, Z algorithm, etc.
Please refer complete article on Check if a string is substring of another for more details!